5t^2+50t-20=0

Solution for 5t^2+50t-20=0 equation:

5t^2+50t-20=0

a = 5; b = 50; c = -20;
Δ = b2-4ac
Δ = 502-4·5·(-20)
Δ = 2900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2900}=\sqrt{100*29}=\sqrt{100}*\sqrt{29}=10\sqrt{29}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-10\sqrt{29}}{2*5}=\frac{-50-10\sqrt{29}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+10\sqrt{29}}{2*5}=\frac{-50+10\sqrt{29}}{10}
$